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Lecture – 06: Linear Momentum Equations

Good morning. Let us start a very interesting subject linear momentum equations which we use for river flow simulations. Today, I will derive linear momentum equations for 3-dimension forms which is the Navier-Stokes equations as well as also derive linear momentum equations for river flow. So, try to compare 2 equations and the complexity of what is there. I will not go to derive step by step. I do encourage all of you to follow any of the fluid mechanics books for getting the step-by step derivation of the Saint-Venanat equations or Navier-Stokes equations, but here I will highlight you the major assumptions, major concept what we follow to derive Navier-Stokes equations as well as Saint-Venanat equations for the river flow. Try to give you the examples and simulations files. Before that, I just want talking about reference book that mostly we are following on the fluvial hydrodynamics book for these presentations, but you can look it the similar type of derivations are available in any others hydraulics book. Let us go for the Euler equations that is what I need to summarize you, last class we discussed that when you consider an infinitely small control volumes which having the dimension of dx dy dz and if you consider the pressures and the velocity both the fields are
continuous functions in these infinite small domains control volume what could be the mass conservation equations. Again we are looking at what could be the linear momentum equations for inviscid flow that
is what we have derived. For inviscid flow, there is no friction components, there is no
viscous stress components. Because of that only you will have the pressure field variations in
the fluid control volumes. So we have only pressure variations. There is no frictional part, no
viscous effects, no shear stress components.
So only we have the pressure variations and considering these control volumes we try to find
out what will be the pressure variations if I am considering these control volumes in a dx/2
distance backward and the forward what could be the pressure at the center point plane the
pressure is p. That is what using a Taylor series expansion of first two terms we can define
the pressure variations and multiplied with area we get the force, pressure multiplied by area
is a force.
So, we know what is the force acting on this plane, similarly what is the force acting in the
back plane of these infinitely small control volumes of dx dy dz using Newton’s second law
of the motions where the sum of the force is acting is equal to the mass into accelerationswhich we know it. So, the same things we have applied for these control volumes as it has the
pressure variabilities and the velocity variations.
(Refer Slide Time: 04:24)
Considering that control volume we got these derivations. If you look at it is very simple
derivation. The pressure force and equating with a gravity force component that what is equal
to the mass into acceleration. That the accelerations component it has local acceleration and
the convective accelerations. More detail about the local acceleration and convective
acceleration you can follow any fluid mechanics books, in which fluid kinematics chapter
you can follow.
You can try to understand it how interesting to know it what is local acceleration? What is the
convective acceleration? That is what is equal to the force due to the pressure in balance that
is what per unit mass that is the reasons we have this part, then you have a gravity force
component. Since it is a three-dimensional, three-coordinate directions, we can write this the
same newtons second law.
So, we can have 3 linear momentum equations, only it is difference between these is that first
one for the x direction and then this we have y direction and you have a z direction and if you
look at this u, v, w, those things are changing and the pressure gradient in, this is the x
direction, this is y direction, this is z direction, acceleration component at x direction, y
direction, z direction.A simple force i.e. mass into acceleration that is what is we have applied it, force is equal to
mass into acceleration at a smaller infinite small control volume where we consider the
pressures field have the continuous functions, the velocity has a continuous functions in terms
of the x, y, z and the t, the space coordinates and the time coordinates. So, considering that
we can develop these equations and this equation is known as Euler equation.
But try to look at these equations these are non-linear partial differential equations. This is the
non-linear partial differential equation in terms of u, v, w and the phi. So, 4 unknowns we
have, so we need to need another equation which is the continuity equation, it is available to
us in a 3-dimensional form. So with these 3 equations plus the continuity equations, the set of
the equations you can solve it, today’s world it is possible using the numerical solutions.
If we solve it, we can get u, v, w and the p variations of any fluid flow problems. That is with
an assumption of inviscid flow, the flow regions where the viscosity does not dominate much
or the significancy of the viscous stresses are very, very less which can be neglected for a
fluid domain, we can apply this Euler equation solvers with having these 3 linear momentum
equations and the continuity equations.
That is what we do it the solutions of the Euler equations, but it is a non-linear partial
differential equation with having 4 dependent variables u, v, w and p and 4 independent
variable x, y, z that is the space coordinates and the t is the time dimension. That is what we
do it and nowadays it is easy to solve these Euler equations with a continuity equation to get
the pressure field as well as the velocity field.
(Refer Slide Time: 08:33)Let us go for the same things, we will extend it to next level. So let us go for the NavierStokes equation derivations. As I said it earlier, I am not going for step by step derivations
which you can get in any fluid mechanics books. Basic my idea is that what the assumptions
we have are and how we are considering it and how we derive the Navier-Stokes equations
with these assumptions for the viscous flow.
So, we are now moving from the Euler equations to Navier-Stokes equation, Euler equation
does not consider the viscosity part whereas Navier-Stokes equations we consider the
viscosity parts. That means in a fluid domain you consider a very, very tiny size of control
volumes which is infinitely small control volumes in which you consider it not only the
pressure and the velocity, the field variation, continuous functions, also we introduce a stress
component.
We consider the control volumes and over this control volume surface we define the stress
components. That is what we follow it, the shear stress component in z direction and the
normal stress. Now, on the surface if you look at this x y z plane, so I have the plane surface
and in this plane surface normal components given σx means it is a plane the perpendicular to
the x direction that are normal stress what I am getting it here.
Normal stress in the x direction over this plane that is what it is this plane having dy and the
dimensions in z is dz, dimensions is the dy dz in this plane which is perpendicular to x that
the normal stress I defined as σx. How do I define the shear stress components? Shear stresscomponents are defined like τxy. What does it indicates here? τxy indicates that it is on the
surface which is the perpendicular to x plane.
The perpendicular to the x plane, the first coordinate is perpendicular to the x plane, on that
surface it is that. The second subscript indicates for us in which direction it acts, in this case it
acts in the y direction. So this is τxy. It is acting on the surface which is perpendicular to x
axis and that axis part what we are looking at that which direction they were.
It can work in z because it is a perpendicular to x plane that can work in the z direction or the
y directions. If it is acting in the y direction, we define it as τxy. If it is acting in the z
direction, we define it as τxz. Please draw a control volume and designate it all these shear
stress components that is what is always difficult for students, but please try to do it that draw
a small infinite small control volume dx dy dz.
Then scale all these normal stress and the shear stress components like here τxy, this is τxz.
Same way if you are coming to this plane if you look at that I am coming to this plane which
is perpendicular to the y direction. This plane perpendicular is in the y direction, if it is that I
have a σy, I have a τyx because this component acting along the x direction, I have τyz.
Same way I can notate it in each plane we can define the stress components. That is what I
am encouraging as a student you please draw the control volumes infinitely small control
volumes with sides dx dy dz and considering these notations please sketch all these stress
components acting on these all the surfaces. Please visualize that. That means you have to
consider control volumes with having 6 faces and each face we want to derive it.
We have to define what is a stress component acting on it, stress into the area is a force. So
that is what we are looking at we have to derive as a stress component and we are designating
with 9 stress components of σx, τxy, τxz and σy, τyx, τyz and σz, τzx, τzy so that is what is called
tensors or stress matrix. So, any small infinitely small control volumes if you consider dx dy
dz and that what we can define as a stress component on the all these 6 faces
All the 6 surfaces or faces we can define this shear stress component. This is valid for the
solid mechanics, this is valid for the fluid mechanics. So, any small solid object if you take it
you can define as a stress field like this or stress tensor like this with having 9 components,but in some of the cases we can take a moment about the axis through the center of the
element and I can write this part which indicates is that just I am equating the moment.
If I going through the axis which is the center of the elements I can get force multiplied by
any distance, force into distance if I equate this part I am getting τxy = τyx. The same concept I
can use it for other faces to find out that τyz = τzy and τzx= τxz .So, that means now instead of 9
components of matrix of shear stress, we can consider it only 6 because other 3 are equal to
that.
We have now only unknowns of 6 stress components that is what is a simplification, which is
what we do it for this infinitely small control volume.
(Refer Slide Time: 15:44)
Now if you understand this concept, then it is very easy to derive this Navier-Stokes
equation, so because we have considered infinitely small control volumes over that we have
defined the stress field. Now we have to define it the stress fields what if we have defined in
the 6 components, they also have a continuous function within that. So again, we can follow
the Taylor series approximations for different faces what could be the value if at the centroid
we know it.
So, that is the reasons what we do it like for example if you look at these figures that I am
looking the σx variations at distance on this surface
Then and is the approximation from Taylor seriesExactly the same way as we did for u and v components, we are doing same approximations
as a σx as a function variability in the space and the time that can be defined as what could be
the value if I know the centroid value, what will be the value at positive side or in the
negative side. Just that means I can know it what are the force components are acting on this
x directions.
Some are the normal stress component as well as the shear stress components. This is a
normal stress multiplied by area, is a force because of normal stress what is the force is acting
it, this is the shear stress along this x direction, how it is acting and also the shear stress along
this x direction because of the z field how it is coming and then you have the force
component because of gravity on these infinitely small control volumes.
That way in ρdVax, which is mass multiplied by acceleration, dV is volume of the control
volume which is equal to dx dy dz. Let me summarize what we have done it now for the same
small control volumes we have trying to write or trying to apply the newtons second laws for
that control volumes where the stress distributions we are considering and from that stress
this one we are trying to find out what is the force acting along the x direction.
That is what we are doing, that is what will be the mass into accelerations in the x direction
that is what we have done it, simple things, the same Newton’s second law of motions,
nothing else, but we have applied for infinitely small control volumes with stress distributions
which is representing a force acting on this surface and also the gravity force as you know it.
So, if you look at that if I simplify these equations, I will get this part.
So that means again we are coming back to the equations which will have this local
accelerations and convective acceleration terms, then we have the terms in terms of σx, τxy, τxz
along the x direction. Same way, looking these equations you can write what could be the
equations for the y direction and the z direction, it is a simple thing. Just look at this equation,
you can write it.
If I try to write the equation motions on along the y axis instead of u I can use the v, instead
of gx I will use gy and similar way corresponding stress components I can write. Once I write
the y axis things, I can write for z axis because all are the similar, only we have changing thescalar velocity field and gx, gy, gz components and the stress components. So the basic
concept is just to apply newtons second law into these smaller control volumes.
Where we have defined the stress variabilities, stress continuum and on that basis we have
derived these components. So, try to understand this and I am encouraging you to write the
equations so that you can feel that what it is coming in.
(Refer Slide Time: 20:38)
Now if you look at that still we have written the equations of motions in terms of
accelerations components, in terms of the stress component, we have not written in terms of
velocity gradients. So, to reduce the number of dependent variables we have to write the
stress in terms of velocity gradient. As we know from Newtons laws of viscosities that the
shear stress is having a linear proportionality to the velocity gradient or the shear strain rate
for Newtonian fluid, the fluid which obeys that laws.
So if you look at that part that means again I can derive the stress components that I have in
terms of velocity gradient that is what is Newton’s laws of the viscosity. So, if you look at
that part that is what is saying that this normal and the shear stress components we can relate
it to the velocity gradient. Viscous stress are proportional to the shear strain rate which is
there in Newtons laws of viscosities.
If you look at that now what is that we have the problem, the problem is not a 1-dimensional
problem, this is a 3-dimensional control volume we have considered it. It will have the 2
deformations, one is a linear deformations another will be volumetric deformations. So weneed to consider the stress due to the linear deformations and the stress due to the volumetric
deformation.
That is what we consider the linear deformations by the dynamic viscosity and the second
viscosity we just introduce it, it is not a dynamic viscosity, it is another viscosity which
consider the stress formations because of volumetric deformations. Thus, any object when
you have the stress field, it will go through linear deformations as well as volumetric
deformation. If you consider that things, I am not going more details.
The normal stress in x, y and z directions can be written in terms of pressure, in terms of
velocity gradient, in terms of volumetric deformations into the µs, µs stands for secondary
viscosity. Please go through advanced fluid mechanics book to try to understand why do we
have the two factors, why do you have a plus and minus, but try to write, try to have a
visualization. Now I can derive the normal stress in the x direction is a function of the p,
which is the pressure.
The velocity gradient representing the linear deformations and I have the volumetric
deformations that is what I can write. Same way I can write for σx, σy and σz. So normal
stress components we have written in terms of p, in terms of velocity gradient which have a
proportionality constants of µ is a dynamic viscosity and µs is the second viscosity which
takes care of volumetric deformations, the stress generations due to the volumetric
deformations.
(Refer Slide Time: 24:21)If you are considering that and go for the next level looking that what could be happened to
the shear stress component that is what any fluid kinematic chapters if you go through we can
easily write it the shear stress component in terms of the velocity gradient with a µ stands for
a dynamic viscosity, so we can write it. Then the Stokes hypothesis says that the second
viscosity effect is a small, that is not big because your volumetric deformations will not have
that much significance.
We can approximate it by that is we can approximate it and similarly with the pressure we
can quantify as the average normal stresses and negative indicates it acts the opposite
directions of the stress field. So, if you look it that way you have
P = 1/3 (σx + σy + σz)
Just trying to locate that if my infinite smaller control volume is tending to the 0 that is what
will be the average, normal stress is equal to the pressures and the directions to take care of
we have introduced the minus. So, now if you look at that if I put it all the stress components
into the basic linear equations of motions, I will get same concept, that means I will get local
accelerations, convective accelerations, the pressure terms are given.
Laplace terms are coming it here plus I have a component because of volumetric
deformation. That is substituting all the variables and which as I said it earlier I am not going
to do a step by step derivations, please look at the step by step variations in any of the books,
but let us understand this equations which is the Navier-Stokes equation. Navier-Stokes
equations for any fluid flow we can solve it if you look it, but what is there in this equation?
If you look at that it is exactly whatever we have this if I consider dynamic viscosity, the
kinematic viscosity is 0 because the non-viscous flow systems it comes out to be Euler
equations that is supposed to be. But that means any Euler equations we have accelerations
component, we have the component of gravity force component, we have the force per unit
area, pressure on balance but we have introduced now the viscous effect that is what you try
to understand.
This is what we have considered the viscous effect as the Laplace equations of u and this part,
it is a volumetric part. So if you try to understand this, this is the part what we haveintroduced in the Navier-Stokes equations, additional components to the Euler equations as
we are considering the fluid has viscosities and that frictional things if I consider as a linear
deformations, as a volumetric deformations, we can make this additional components.
Any of the computational fluid dynamics that they in very detail discuss about these
equations, but I just want to summarize it what we are trying to get, we are again getting a
non-linear partial differential equation with a more nonlinear component like Laplace second
order components are there. So, if you look at this way, the more complicated equations we
are getting.
It as nonlinear partial differential equations to solve the fluid flow problems considering the
viscous effects and here also we have this 4 dependent variables, what are they? They are
velocity u, v, w, velocity scalar components and the pressures but we have the 4 equation, 3
linear momentum equations and 1 continuity equation. So, if considering that we can solve it
and today it is possible to solve it.
There lot of solvers are there for the Navier-Stokes equations commercially or many sources
are there nowadays you can get a Navier-Stokes equation solvers that is not a big issue to
solve these equations, but it looks a very complicated equations in terms of non-linear partial
differential equations with 4 dependent variable u, v, w and p, 4 independent variable the
space coordinates like x, y, z and the time. So, how to solve these things I am not going to tell
most things now.
(Refer Slide Time: 29:42)Let us come back to just solve these equations taking example problems which is we are
considered from FM White Fluid Mechanics text book just to demonstrate it that we can use
this equation to get some analytical solutions for a fluid flow problems. Let us consider the
flow is incompressible, density does not vary with time, and velocity field is given.
And if you look at this velocity field you can interpret many things like the velocity field
does not have the time components, it has only x and y, there is no z component and the w
part is equal to 0, so this have to consider it before solving the problems because when you
try to get analytical solutions you try to understand what type of velocity field is given to us
and the density field. Density says that is incompressible that means density is not varying.
So, if you are consider that part we have to determine under what conditions, that means we
are looking the solutions of the Navier-Stokes equations. If this is the velocity field what is
the pressure field we are looking at or does this velocity field satisfy the Navier-Stokes
equations that is what we are looking at. So, what we do it like we just try to apply this
Navier-Stokes equation and substitutes all the terms here.
All these terms we just compute it in Navier-Stokes equations of the first x and y and z
directions. substitutes all these things, this is what we do it first partial derivative, second
partial derivative with respect to x, with respect to the y, then we substitute these values and
we try to find out that what will be these 3 equations.
If you look at these 3 equations very good you can find out these equations are very simple
because we can integrate it. It does not have any component, so we can integrate this,
integrations with this we can get it next part.
(Refer Slide Time: 32:11)So, that if we integrate it you can see this pressure is varying like this, but there is a constant
which may be functions of x and y, we do not know it. Second point is coming okay we have
integrated one equation to get it what could be probable solutions of this with an unknown
function f1(x), is it correct one? So what do we do it again we use this equations to check it
other two equation do they satisfy ?
That is what we did it, differentiate with respect to y respect to x, then you just compare it
and we found it these are also the same values. That means this velocity field what is given
satisfies the Navier-Stokes equations. Now we are looking at what could be the pressure field.
(Refer Slide Time: 33:06)
Again, these are all I can say that is just differences in integrations to just find out that what
will be the integration constant values. That is the reasons again we substitute equation 4 intoequation 1, 2 into the 1 and obtain what will be the f1(x). So, once you know this f1(x), then
we integrate it and do the differentiations still we get the f2(y) value.
(Refer Slide Time: 33:39)
Now we will substitute it which gives us the pressure field. This part we got it. This is the
part we got it after integration and differentiation of the previous equations and that is what is
giving us the pressure field. We satisfy the velocity field, pressure field for this the flow
domains where the Navier-Stokes equation holds good.
(Refer Slide Time: 34:08)
Before completing these things, I just want to show you that a similar way we have applied
the Navier-Stokes equations for the fluid flow problems in channels, we know it many of the
times we put a small obstruction like a porcupine structures, the river training works and try
to know it does it work?, how much it is working in terms of velocity reductions. That is whatwe try to do with 3-dimensional models, CCHE 3D, 3-dimensions models which has the
Navier-Stokes equations with depth integrated.
The turbulent flow which will come later on. We will have a Eddy viscosity, k-€ model and
the depth integrated logarithmic law and all and it is a solver of the finite element approach
that solutions what gives is just going to give you confidence that we have put the structures
here, we try to measure the velocities using the numerical models.
(Refer Slide Time: 34:58)
And if you look at this velocity distributions obtained from flume experiments. You can see
that there is a velocity reduction because of these obstructions that is what is clearly visible
from upstream and the downstream velocity distributions. There are significant velocity
reductions are happening in experimental data as well as the numerical modeling of NavierStokes equations.
So, now it is possible, there are the models available which solves the Navier-Stokes
equations with some approximations like depth integrated concept with turbulent structures,
we can get the solutions like what we have done it for a river training work having the
porcupine structures, a small obstruction we can see that how much velocity reductions are
happening.
Same things you can look at how much velocity reductions happening at the experimental
levels. So, those things we can conduct and it is possible nowadays we can do it as we
consider as Navier-Stokes equation solvers.(Refer Slide Time: 36:23)
Let us come back to derive the momentum equations for river flow, unsteady gradually
varied flow. The variations in the gradually as well as unsteady flow. If you look at a river as
we have considered here is the river, this is the bed, this is the free surface and there will be
the energy gradient line. This is a hydraulic gradient line and if I consider these channels
having making angle ϴ.
The channel is making or the river is making angle ϴ and this is the free surface, this is the
energy gradient line, the slope of energy gradient line we define as Sf and we also define the
slope of bed is S0 which is equal to tanϴ. But this cross-section geometry of the rivers can
have a complex like this as we are looking for area integrated concept, it can have a like this,
the free surface can be here and Zc is the centroid point where the pressure acts it.
So, in this field p is having the variations, the pressures acting on this field is having the
variations, A also varies. So p multiplied by A also have the variabilities, pressure into area is
the force, force also we are considering is a variability from along the x direction. Along the
x direction, the force into area varies. If this is my control volume, I can easily interpret it
what are the forces acting on this control surface.
One is the force components hydrostatic or hydrodynamic part, here is hydrodynamic force
components are there acting on these two surfaces, the force acting on the free surface we can
negligible it but there will be a force, the frictional force acting on this. The frictional forceacting on the bed surface we can quantify as τo P dx if you look at this τo stands for the bed
shear stress.
The shear force acting on the bed times of the perimeter of the flow, weighted perimeter of
the flow and to the dx coordinates. That is what the frictional force acting on this. Frictional
force we have designated as a bed shear stress acting on this bed that what we are defining it.
Then we are defining the gravity force components. As it is an inclined phase you can find
out it has a component in x directions, along this direction.
Along the x direction as a gravity component of Fw sinϴ and Fw cos ϴ components. So that
resolving we have done it, that is what is the bed frictional resistance is τo times of wetted
perimeter and length. Fw is the weight of the fluid in this control volume, so we have
considered it and ϴ is inclusive. Now again we have to apply the same Newtons second law
okay. Find out the net force difference in the x directions that is a pA and the force acting on
this part.
Then the force component due to the gravity, frictional force is equal to mass into
acceleration. If you look at that is the same concept we are assuming it, only we have
changed the control volume, we have changed the force components. We have a different
assumption for that like in a free surface there is no force component. We have considered
this hydrodynamic force is a pA which is variability we have considered it.
And we find out the frictional resistance force and all that is what is some of the forces we
did it is equal to mass into acceleration that is the component. And the accelerations we can
define as only this one directions area average U value that is what we can define as local
velocity and the convective velocity in the x direction. So, we have a one-dimensional
equation in x direction.
Let us try to conceptualize that we have considered the river and we are just looking at the
longitudinal directions what is happening, we are not considering much lateral direction what
is happening or the depth variations what is. We are just looking at along the longitudinal
directions how these force components are there that is what is the area average velocity we
have, we are looking at the local accelerations, we are looking the convective acceleration
terms.(Refer Slide Time: 41:28)
Now we have to simplify these things the equations components. So if you look at this part,
we